Theorem sbi1 2380

Description: Removal of implication from substitution. (Contributed by NM, 14-May-1993.)

Assertion

Ref	Expression
sbi1	⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

Proof of Theorem sbi1

Step	Hyp	Ref	Expression
1		sbequ2 1869	. . . . 5 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → 𝜑))
2		sbequ2 1869	. . . . 5 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑 → 𝜓) → (𝜑 → 𝜓)))
3	1, 2	syl5d 71	. . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → 𝜓)))
4		sbequ1 2096	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜓 → [𝑦 / 𝑥]𝜓))
5	3, 4	syl6d 73	. . 3 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
6	5	sps 2043	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
7		sb4 2344	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
8		sb4 2344	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑 → 𝜓) → ∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓))))
9		ax-2 7	. . . . . 6 ⊢ ((𝑥 = 𝑦 → (𝜑 → 𝜓)) → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜓)))
10	9	al2imi 1733	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜓)))
11		sb2 2340	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) → [𝑦 / 𝑥]𝜓)
12	10, 11	syl6 34	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜑 → 𝜓)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜓))
13	8, 12	syl6 34	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑 → 𝜓) → (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜓)))
14	7, 13	syl5d 71	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
15	6, 14	pm2.61i 175	1 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1473  [wsb 1867

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-12 2034  ax-13 2234

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868

This theorem is referenced by:  spsbim  2382  sbim  2383  2sb5ndVD  38168  2sb5ndALT  38190